How could I prove that $x^2+1$ and $x^4+1$ are irreductible over $\mathbb{Z}/2 \mathbb{Z}$ and $\mathbb{Z}/3 \mathbb{Z}$?

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) (\mod I)$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreductible in $A[x]$

I think I have to use this theorem, but I am not certain. Is anyone could help me at this point?


marked as duplicate by Jyrki Lahtonen abstract-algebra Mar 9 '16 at 8:40

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  • $\begingroup$ Maybe I'm crazy, but I think these are reducible, at least in $\mathbb Z/2 \mathbb Z$. You can factor them into linear terms. $\endgroup$ – Alfred Yerger Mar 9 '16 at 3:33
  • $\begingroup$ They first is not irreducible over $\mathbb Z/2\mathbb Z$ $\endgroup$ – Thomas Andrews Mar 9 '16 at 3:33
  • $\begingroup$ Neither of those polynomials are irreducible in $\mathbb Z/2\mathbb 2Z$. A good first step before checking irreducibility is to see whether the polynomials have roots! In your case there are only finitely ($2$ and $3$ respectively) many candidates to check. $\endgroup$ – Ravi Mar 9 '16 at 3:33
  • $\begingroup$ What about $\mathbb{Z}/3\mathbb{Z}$? I would like to factorise $x^8-1$ in product of irreductibles in the rings $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$? The problem is I really don't how to start. $\endgroup$ – Taj Mohamed Bandalandabad Mar 9 '16 at 3:36
  • $\begingroup$ If you don't know how to start you can SEARCH for similar questions. Either here or in your lecture notes/textbook. You SHOULD NOT ASK THE SAME QUESTION MANY TIMES: $\endgroup$ – Jyrki Lahtonen Mar 9 '16 at 8:41

.$x^2+1=x^2+1^2=(x+1)^2$ in $2\mathbb{Z}$

$x^4+1 = (x^2+x+2)(x^2+2x+2)$ in $3\mathbb{Z}$

Neither is irreducible mod it's respective field. In fact, $x^4+1$ is reducible mod $n\mathbb{Z}$ for every $n$!


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